Right option is (a) 2 cos (\(\frac{2π}{t}\) kt0) X[k]
For explanation: x (t-t0) is periodic with period T. the Fourier series coefficient of x (t-t0) is X1[k] = \(\frac{1}{T}\) ∫ x (t-t0)e^-jkωt dt
= e^-jkωt0 X[k]
Similarly, the Fourier series coefficient of x (t+t0) is X2[k] = e^jkωt0 X[k]
The Fourier series coefficient of x (t-t0) + x (t+t0) is
Y[k] = X1[k] + X2[k]
= e^-jkωt0 X[k] + e^jkωt0 X[k]
= 2 cos (\(\frac{2π}{t}\) kt0) X[k].